Question: Divide the following complex numbers. $ \dfrac{-9-6i}{-3-2i}$
Explanation: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-3+2i}$ $ \dfrac{-9-6i}{-3-2i} = \dfrac{-9-6i}{-3-2i} \cdot \dfrac{{-3+2i}}{{-3+2i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-9-6i) \cdot (-3+2i)} {(-3-2i) \cdot (-3+2i)} = \dfrac{(-9-6i) \cdot (-3+2i)} {(-3)^2 - (-2i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-9-6i) \cdot (-3+2i)} {(-3)^2 - (-2i)^2} = $ $ \dfrac{(-9-6i) \cdot (-3+2i)} {9 + 4} = $ $ \dfrac{(-9-6i) \cdot (-3+2i)} {13} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-9-6i}) \cdot ({-3+2i})} {13} = $ $ \dfrac{{-9} \cdot {(-3)} + {-6} \cdot {(-3) i} + {-9} \cdot {2 i} + {-6} \cdot {2 i^2}} {13} $ Evaluate each product of two numbers. $ \dfrac{27 + 18i - 18i - 12 i^2} {13} $ Finally, simplify the fraction. $ \dfrac{27 + 18i - 18i + 12} {13} = \dfrac{39 + 0i} {13} = 3 $